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Modern Maths

Probability

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CAT 2025 Lesson : Probability - Mutually Exclusive Events

3. Coins, Dice and Cards

In a lot of entrance tests, we are expected to understand tossing a coin, rolling a die (or dice) or picking a card from a standard deck of cards. Sometimes the questions do not explain these events in detail. Therefore, we have explained these below.

1

) Tossing a coin: A standard coin has two sides to it. One is called Heads and the other is called Tails.
In an unbiased coin, these

\bm{2}

outcomes are possible and both of them are equally likely.

2

) Rolling a die: A standard die is a cube with

6

sides. Each side is uniquely numbered –

\bm{1, 2, 3, 4, 5}

and

\bm{6}

. These

6

outcomes are equally likely when a die is rolled. (Note that dice is the plural for die.)

3

) Standard deck of cards: There are a total of

52

cards in a standard deck.
Ranks: There are

13

ranks.

9

of them are numbered from

2

10

, followed by Jack (J),
Queen (Q), King (K) and Ace (A).
The

13

cards stated in order are

-

2, 3, 4, 5, 6, 7, 8, 9, 10

, J, Q, K
Suits: There are

4

suits – Clubs, Diamonds, Hearts and Spades. Each of these suits contain the

13

differently-ranked cards mentioned above. Therefore, a deck contains

4 \times 13 = 52

cards
Colour: Each of the cards are either Red or Black. The colour is based on the suit of the card. Clubs and Spades are black-coloured cards while Diamonds and Hearts are red-coloured cards.
Therefore, there are

26

black cards and

26

red cards in a standard deck.

4. Exclusivity of Events

4.1 Mutually Exclusive Events

When

2

events are such that at most one of them can occur at a time and not both, they are called mutually exclusive events. For instance, when we toss a coin, getting a head or a tail are mutually exclusive events.

In other words, two or more events are mutually exclusive if there is no common outcome between them, i.e., P(A

\cap

= \bm{0}

Probability of occurrence of either of the two mutually exclusive events (A or B) is P(A or B)

=

P(A

\cup

=

P(A) + P(B)

Example 4

There are

3

black balls,

4

red balls and

6

white balls. What is the probability of drawing a black or red ball?

(

1

)

\dfrac{3}{13}

(

2

)

\dfrac{6}{13}

(3)

\dfrac{7}{13}

(4)

\dfrac{9}{13}

Solution

Total number of outcomes

= n(\text{S}) = 3 + 4 + 6 = 13

No. of favourable outcomes

= n(\text{E}) = 3 + 4 = 7

P(E)

= \dfrac{n(\text{E})}{n\text(S)} = \dfrac{7}{13}

Answer: (

3

)

\dfrac{7}{13}

Example 5

Four coins are tossed simultaneously. What is the probability of
(I) getting a head in at least two of the coins?
(II) getting a head in at least one of the coins?

Solution

Case I: Each coin has

2

outcomes – head or a tail. Imagine the

4

coins being tossed
simultaneously by

4

people standing in a row. Each of these coins can have

2

outcomes.
Total outcomes when

4

coins are tossed

= 2 \times 2 \times 2 \times 2 = 16

Favourable outcomes are

2

heads &

2

tails,

3

heads &

1

tails and

4

heads & no tails.
Let H denote heads and T denote tails. Then, HHTT signifies the results of the coins in that order – coins of the first two people returning heads and the last two returning tails. Number of different such outcomes with

2

heads and

2

tails is the number of ways letters in the word HHTT can be permuted.

Number of permutations for HHTT

= \dfrac{4 !}{2 ! \times 2 !} = 6

; HHHT

= \dfrac{4 !}{3 !} = 4

; HHHH

= \dfrac{4 !}{4 !} = 1

Probability

= \dfrac{6 + 4 + 1}{16} = \dfrac{\bm{11}}{\bm{16}}

Case II: We shall first find the probability of getting a tail in all coins and then subtract this from

1

.

Number of permutations for TTTT

= \dfrac{4 !}{4 !} = 1

Probability of getting TTTT

= \dfrac{1}{16}

Probability of getting at least

1

head

= 1 - \dfrac{1}{16} = \dfrac{\bm{15}}{\bm{16}}

Answer: (I)

\dfrac{11}{16}

; (II)

\dfrac{15}{16}

Example 6

When two six-sided dice are rolled, what is the probability of the sum of the numbers on the two dice being greater than

9

Solution

As there are two dice and each of which have

6

outcomes,
Total outcomes when two dice are rolled

= 6 \times 6 = 36

Favourable outcomes are when the sum of the number are

10, 11

12

10

⇒ {

(6, 4), (5, 5), (4, 6)

}

11

⇒ {

(6, 5), (5, 6)

}

12

⇒ {

(6, 6)

}

No. of favourable outcomes

= 6

Probability

= \dfrac{6}{36} = \dfrac{1}{6}

Answer:

\dfrac{1}{6}

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